Bound state

A bound state is a composite of two or more fundamental building blocks, such as particles, atoms, or bodies, that behaves as a single object and in which energy is required to split them.^[1]

In

In relativistic quantum field theory, a stable bound state of n particles with masses ${\displaystyle \{m_{k}\}_{k=1}^{n}}$ corresponds to a

${\displaystyle \textstyle \sum _{k}m_{k}}$

• A proton and an electron can move separately; when they do, the total center-of-mass energy is positive, and such a pair of particles can be described as an ionized atom. Once the electron starts to "orbit" the proton, the energy becomes negative, and a bound state – namely the hydrogen atom – is formed. Only the lowest-energy bound state, the ground state, is stable. Other excited states are unstable and will decay into stable (but not other unstable) bound states with less energy by emitting a photon.
• A positronium "atom" is an unstable bound state of an electron and a positron. It decays into photons.
• Any state in the quantum harmonic oscillator is bound, but has positive energy. Note that ${\displaystyle \lim _{x\to \pm \infty }{V_{\text{QHO}}(x)}=\infty }$ , so the below does not apply.
• A nucleus is a bound state of protons and neutrons (nucleons).
• The proton itself is a bound state of three quarks (two up and one down; one red, one green and one blue). However, unlike the case of the hydrogen atom, the individual quarks can never be isolated. See confinement.
• The
  atoms can form a bound pair in an optical lattice.^[5][6][7] The JCH Hamiltonian also supports two-polariton bound states when the photon-atom interaction is sufficiently strong.^[8]

Let σ-finite measure space ${\displaystyle (X,{\mathcal {A}},\mu )}$ be a probability space associated with separable complex Hilbert space ${\displaystyle H}$. Define a one-parameter group of unitary operators ${\displaystyle (U_{t})_{t\in \mathbb {R} }}$, a

${\displaystyle \rho =\rho (t_{0})}$

${\displaystyle T}$

${\displaystyle H}$

${\displaystyle \mu (T,\rho )}$

${\displaystyle T}$

${\displaystyle \rho }$

${\displaystyle \rho (t_{0})\mapsto [U_{t}(\rho )](t_{0})=\rho (t_{0}+t)}$

is bound with respect to ${\displaystyle T}$ if

${\displaystyle \lim _{R\rightarrow \infty }{\sup _{t\geq t_{0}}{\mu (T,\rho (t))(\mathbb {R} _{>R})}}=0}$,

where ${\displaystyle \mathbb {R} _{>R}=\lbrace x\in \mathbb {R} \mid x>R\rbrace }$.^{[dubious – discuss][9]}

A quantum particle is in a bound state if at no point in time it is found “too far away" from any finite region ${\displaystyle R\subset X}$. Using a wave function representation, for example, this means

${\displaystyle {\begin{aligned}0&=\lim _{R\to \infty }{\mathbb {P} ({\text{particle measured inside }}X\setminus R)}\\&=\lim _{R\to \infty }{\int _{X\setminus R}|\psi (x)|^{2}\,d\mu (x)},\end{aligned}}}$

such that

${\displaystyle \int _{X}{|\psi (x)|^{2}\,d\mu (x)}<\infty .}$

In general, a quantum state is a bound state if and only if it is finitely normalizable for all times ${\displaystyle t\in \mathbb {R} }$.^[10] Furthermore, a bound state lies within the pure point part of the spectrum of ${\displaystyle T}$ if and only if it is an eigenstate of ${\displaystyle T}$.^[11]

More informally, "boundedness" results foremost from the choice of

For a concrete example: let ${\displaystyle H:=L^{2}(\mathbb {R} )}$ and let ${\displaystyle T}$ be the

${\displaystyle \rho =\rho (0)\in H}$

${\displaystyle [-1,1]\subseteq \mathrm {Supp} (\rho )}$

• If the state evolution of ${\displaystyle \rho }$ "moves this wave package to the right", e.g. if ${\displaystyle [t-1,t+1]\in \mathrm {Supp} (\rho (t))}$ for all ${\displaystyle t\geq 0}$, then ${\displaystyle \rho }$ is not bound state with respect to position.
• If ${\displaystyle \rho }$ does not change in time, i.e. ${\displaystyle \rho (t)=\rho }$ for all ${\displaystyle t\geq 0}$, then ${\displaystyle \rho }$ is bound with respect to position.
• More generally: If the state evolution of ${\displaystyle \rho }$ "just moves ${\displaystyle \rho }$ inside a bounded domain", then ${\displaystyle \rho }$ is bound with respect to position.

Properties

As finitely normalizable states must lie within the

so that ψ is exponentially suppressed at large x. This behaviour is well-studied for smoothly varying potentials in the WKB approximation for wavefunction, where an oscillatory behaviour is observed if the right hand side of the equation is negative and growing/decaying behaviour if it is positive.^[14] Hence, negative energy-states are bound if V vanishes at infinity.

Non-Degeneracy in One dimensional bound states

1D bound states can be shown to be non degenerate in energy for well-behaved wavefunctions that decay to zero at infinities. This need not hold true for wavefunction in higher dimensions. Due to the property of non-degenerate states, one dimensional bound states can always be expressed as real wavefunctions.

Proof

Consider two energy eigenstates states ${\textstyle \Psi _{1}}$ and ${\textstyle \Psi _{2}}$ with same energy eigenvalue. Then since, the Schrodinger equation, which is expressed as: $${\displaystyle E=-{\frac {1}{\Psi _{i}(x,t)}}{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}\Psi _{i}(x,t)}{\partial x^{2}}}+V(x,t)}$$ is satisfied for i = 1 and 2, subtracting the two equations gives: $${\displaystyle {\frac {1}{\Psi _{1}(x,t)}}{\frac {\partial ^{2}\Psi _{1}(x,t)}{\partial x^{2}}}-{\frac {1}{\Psi _{2}(x,t)}}{\frac {\partial ^{2}\Psi _{2}(x,t)}{\partial x^{2}}}=0}$$ which can be rearranged to give the condition: $${\displaystyle {\frac {\partial }{\partial x}}\left({\frac {\partial \Psi _{1}}{\partial x}}\Psi _{2}\right)-{\frac {\partial }{\partial x}}\left({\frac {\partial \Psi _{2}}{\partial x}}\Psi _{1}\right)=0}$$ Since ${\textstyle {\frac {\partial \Psi _{1}}{\partial x}}(x)\Psi _{2}(x)-{\frac {\partial \Psi _{2}}{\partial x}}(x)\Psi _{1}(x)=C}$, taking limit of x going to infinity on both sides, the wavefunctions vanish and gives ${\textstyle C=0}$. Solving for ${\textstyle {\frac {\partial \Psi _{1}}{\partial x}}(x)\Psi _{2}(x)={\frac {\partial \Psi _{2}}{\partial x}}(x)\Psi _{1}(x)}$, we get: ${\textstyle \Psi _{1}(x)=k\Psi _{2}(x)}$ which proves that the energy eigenfunction of a 1D bound state is unique. Furthermore it can be shown that these wavefunctions can always be represented by a completely real wavefunction. Define real functions ${\textstyle \rho _{1}(x)}$ and ${\textstyle \rho _{2}(x)}$ such that ${\textstyle \Psi (x)=\rho _{1}(x)+i\rho _{2}(x)}$. Then, from Schrodinger's equation: $${\displaystyle \Psi ''=-{\frac {2m(E-V(x))}{\hbar ^{2}}}\Psi }$$ we get that, since the terms in the equation are all real values: $${\displaystyle \rho _{i}''=-{\frac {2m(E-V(x))}{\hbar ^{2}}}\rho _{i}}$$ applies for i = 1 and 2. Thus every 1D bound state can be represented by completely real eigenfunctions. Note that real function representation of wavefunctions from this proof applies for all non-degenerate states in general.

Node theorem

Node theorem states that n-th bound wavefunction ordered according to increasing energy has exactly n-1 nodes, ie. points ${\displaystyle x=a}$ where ${\displaystyle \psi (a)=0\neq \psi '(a)}$. Due to the form of Schrödinger's time independent equations, it is not possible for a physical wavefunction to have ${\displaystyle \psi (a)=0=\psi '(a)}$ since it corresponds to ${\displaystyle \psi (x)=0}$ solution.^[15]

Requirements

A boson with mass m_χ mediating a weakly coupled interaction produces an Yukawa-like interaction potential,

${\displaystyle V(r)=\pm {\frac {\alpha _{\chi }}{r}}e^{-{\frac {r}{\lambda \!\!\!{\frac {}{\ }}_{\chi }}}}}$,

where ${\displaystyle \alpha _{\chi }=g^{2}/4\pi }$, g is the gauge coupling constant, and ƛ_i = ℏ/m_ic is the

• Bethe–Salpeter equation
• Bound state in the continuum
• Composite field
• Cooper pair
• Resonance (particle physics)
• Levinson's theorem

Remarks

References